MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Fall 2008
Midterm exam, 7-9pm (120 mins/100 pts)
6.436J/15.085J
10/21/08
Problem 1: (15 points)
Let {Xn } be a sequence of random variables (i.e., measurable functions) defined
on the same probability space (Ω, F, P). Show, from first principles (i.e., only
using the definition of σ-fields and of measurability) that the event {supn Xn <
c} is measurable.
Solution: Observe that the event
sup Xn ≤ c,
n
is measurable for any c because
{sup Xn ≤ c} =
�
n
{Xn ≤ c},
n
and now we can just write
{sup Xn < c} =
n
∞
{sup Xn ≤ c − 1/n},
n=1
n
so the event in question is a countable union of measurable sets.
Problem 2: (15 points)
Let {Xn } and {Yn } be two nondecreasing sequences of random variables, de­
fined on the same probability space (Ω, F, P). (That is, Xn (ω) ≤ Xn+1 (ω), for
all ω, and similarly for Yn .) Let X = limn→∞ Xn and Y = limn→∞ Yn . As­
sume that for every n, the random variables Xn and Yn are independent. Show
(using only the definitions and basic facts about measures) that X and Y are
independent. Hint: Use the definition of independence in terms of CDFs.
Solution: Observe that
{X ≤ x, Y ≤ y} =
∞
�
{Xi ≤ x, Yi ≤ y}.
i=1
Indeed, if ω is in the set on the left-hand side, then it must be in each set onn the
right-hand side since Xi (ω) ≤ X(ω) and Yi (ω) ≤ Y (ω) for all i. Conversely, if
1
ω belongs to each of the sets on the right hand side, then Xi (ω) ≤ x, Yi (ω) ≤ y
for all i, and passing to the limit we get X(ω) ≤ x, Y (ω) ≤ y, so ω must belong
to the set on the left-hand side.
The monotonicity of Xi , Yi implies that {Xi ≤ x, Yi ≤ y} is a decreasing
sequence. So, using the continuity property of probability measures
P(X ≤ x, Y ≤ y) = lim P(Xn ≤ x, Yn ≤ y),
n
and now we use the independence of Xn , Yn to get
P(X ≤ x, Y ≤ y) = lim P(Xn ≤ x)P(Yn ≤ y) = lim P(Xn ≤ x) lim P(Yn ≤ y),
n
n
n
and now using continuity of probabilities again,
P(X ≤ x, Y ≤ y) = P(X ≤ x)P(Y ≤ y),
which is what we needed to show.
Problem 3: (15 points)
Let F1 and F2 be two CDFs, and suppose that F1 (t) < F2 (t), for all t. As­
sume that F1 and F2 are continuous and strictly increasing. Show that there
exist random variables X1 and X2 , with CDFs F1 and F2 , defined on the same
probability space such that X1 > X2 .
Solution: Observe that since F1 (t), F2 (t) are continuous and strictly increasing,
the functions F1−1 (t), F2−1 (t) are well-defined when t ∈ (0, 1). Let U be a
uniform random variable in (0, 1) and let
X1 = F1−1 (U )
X2 = F2−1 (U )
Then the CDF of X1 is F1 since
P(X1 ≤ t) = P(F1−1 (U ) ≤ t) = P(U ≤ F1 (t)) = F1 (t),
and similarly the DF of X2 is F2 . Since F2 > F1 it follows that F2−1 < F1−1
and so X1 > X2 .
Problem 4: (20 points)
Let {Xn } be a sequence
�of independent random variables. Show that supn Xn <
∞, a.s., if and only if ∞
n=1 P(Xn > c) < ∞ for some c.
2
Remark: The problem indicates that Xi are random variables (as opposed to
extended-valued random variables), which means that their range is the real
numbers. On the other hand, we can’t say the same about supi Xi , since it
can take the value of +∞.
�
Solution: Suppose ∞
n=1 P(Xn > c) = ∞ for all c. For any c, the probability
of supi Xi < c must be 0 since the event Xi > c occurs infinitely often with
probability 1 by the Borel-Cantelli lemma. It follows that
�
�
P(sup Xi finite ) = P(∪∞
P(sup Xi < n) =
0 = 0,
n=0 sup Xi < n) ≤
i
i
i
n
n
so supi Xi must equal +∞ with
probability 1.
�∞
Suppose that for some c, n=1 P(Xn > c) < ∞. By the Borel-Cantelli
lemma, this means that with probability 1, the number of times Xi > c is finite.
If Xi > c occurs finitely many times, then supi Xi is finite, and it follows that
supi Xi is finite with probability 1.
Problem 5: (20 points)
Let X be a continuous nonnegative random variable with PDF f . Let N = �X�
be the integer part of X (i.e., N is the largest integer that satisfies N ≤ X). Let
U = X − N .
(a) Write down an expression (in terms of f ) for the conditional PDF fU |N (u |
n) of U , given N .
(b) Write down an expression for the PDF of U .
(c) Write down an expression for the conditional PMF of N given U .
Solution:
(a)
fU |N (u | n) =
fU,N (u, n)
f (u + n)
= � n+1
,
fN (n)
f (x)dx
n
where u ∈ [0, 1), and n is a nonnegative integer.
(b)
fU (u) =
∞
�
fU |N (u|n)fN (n) =
n=0
∞
�
� n+1
n=0
for u ∈ [0, 1).
3
f (u + n)
n
f (x)dx
�
n+1
f (x)dx =
n
∞
�
n=0
f (u+n),
(c)
fN |U (n | u) =
fN,U (n, u)
f (n + u)
=
�
∞
,
fU (u)
k=0 f (k + u)
for n integer, u ∈ [0, 1).
Problem 6: (15 points)
Let X and Y be positive continuous random variables with known joint PDF
fX,Y . Find an expression for the joint PDF of X and XY .
Solution: If U = X, V = XY , then X = U, Y = V /U , and the determinant of
the Jacobian of the inverse transformation is
��
�
�
�
1
0
�� 1
�
�
−v/u2 1/u �
= u
,
so that
fU,V (u, v) =
1
fX,Y (u, v/u),
u
for u, v > 0.
4
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6.436J / 15.085J Fundamentals of Probability
Fall 2008
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